# Intuitive idea of the Lipschitz function

I'm trying to understand intuitively the notion of Lipschitz function.

I can't understand why bounded function doesn't imply Lipschitz function.

I need a counterexample or an intuitive idea to clarify my notion of Lipschitz function.

I need help

Thanks a lot

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also assume continuity? –  Halil Duru Apr 6 '13 at 20:38

## 5 Answers

A Lipschitz function is such that $$|f(x)-f(y)|\leq \alpha |x-y|$$ for any points you pick. Writing this as $$\left|\frac{f(x)-f(y)}{x-y} \right|\leq \alpha$$

what we're saying is that the slope of the secant line joining $(x,f(x))$ and $(y,f(y))$ is always bounded above by $\alpha$. An example of a Lipschitz function is $\sin x$, or $x$. An example of a function which is not Lipschitz but is bounded is $$\sin (x^2)$$ over $\Bbb R$. This is because as we go further towards $+\infty$, the oscillation becomes faster, and thus the slope of the secant lines get nearer and nearer to vertical ones.

An example of a function that is not Lipschitz nor bounded is $\sqrt x$ over $\Bbb R_{>0}$. This is because if we fix $x=0$ and make $y$ very close to $0$, the slope of the secant line grows without bound.

Finally, we can give an example of a function which is Lipschitz but isn't bounded: $x+\sin x$ over $\Bbb R$. Its slope will never get larger than $1$.

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Excellent explanation! +1 –  Clayton Apr 6 '13 at 20:33
If a function has a derivative that is bounded, must it be a Lipschitz function? –  Peter Olson Apr 7 '13 at 0:28
@PeterOlson Assume $f$ has bounded derivative. By the Mean Value Theorem, for each $x,y\in\operatorname{dom}f$, what can you equate $$\frac{f(x)-f(y)}{x-y}$$ to? –  Pedro Tamaroff Apr 7 '13 at 0:34

Look at the square root function on $[0,1]$ and its behaviour at 0.

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Look at the unitary circle, $f(x)=\sqrt{1-x^2}$ how is the tangent line in $x=1$?

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Hint: $f$ differentiable and Lipschitz $\Rightarrow$ $f'$ bounded.

But $f$ bounded and differentiable $\not\Rightarrow$ $f'$ bounded.

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With this version (just assuming boundedness) , any discontinuous bounded function forms a counter-example .[why?]

But if you also assume continuity , see the examples below/above.

Furthermore , you may even have differentiability and boundedness without Lipschitzness......

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How could a function be differentiable without being lipschitz? –  treble Apr 6 '13 at 21:01
Look at Tamaroff's example $sin[x^2]$ ..Isn't it differentiable? –  Halil Duru Apr 6 '13 at 21:19
OK, I wasn't thinking about the derivative going to infinity. –  treble Apr 6 '13 at 22:28
Exactly.......... –  Halil Duru Apr 6 '13 at 22:29